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AC 20161207 exercise 1
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子字串
請設計一個程式，使用者會輸入兩個字串，兩個字串間用一個空格隔開，每個字串的內容只會包含大小寫的英文字母，請判斷第二個字串是否出現在第一個字串內。請將大小寫視為同一個字。
輸入範例1：
friEND end
輸出範例1：
There is "end" in "friEND"
輸入範例2：
friEND friends
輸出範例2：
There are no "friends" in "friEND"
asked
Dec 7, 2016
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2016-1 程式設計(一)AC
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Shun-Po
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answered
Dec 18, 2016
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410523016
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answered
Dec 18, 2016
by
410523016
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Hidden content!#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
int main()
{
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answered
Dec 18, 2016
by
Gulang
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-122
points)
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Hidden content!#include <stdio.h>
#include * * *
#include <ctype.h>
int main()
{
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answered
Dec 18, 2016
by
Gulang
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-122
points)
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
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{
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*** ** ** * * * * * ** 0;
}
answered
Dec 18, 2016
by
Jim870729
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
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* * * * * *** * *** * ** 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
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Hidden content!#include<stdio.h>
* ** ** *
#include<ctype.h>
int main ( void )
{
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* * * * * **** * * * 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
* * * * * * * ** * i,j;
** * **** * * *** * ** * *** * * * * * * *
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** * * * ** *** * * * * ** * *** * * * ** * ** ** * ** is %s in %s",b,a);
** * * * ** ** * ** *** * * 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
points)
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
* * * **** * * ** **** i,j;
* *** ** * * ** * ** ***** * * *
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** * * * *** * * * * *
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* **** * * * *** ** ** ** ** *
* * * ** *** * * * ** = strstr(a,b);
***** * ** ** * * * ***
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* * * * ** ** **
* ** * ** * ** * ** * ** ** * * * ** ** * * **** * * **** is %s in %s",b,a);
* ** **** * * * ** * * ** ** 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
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Hidden content!#include * * *** **** **
#include * * * ** *
#include * *** * **** *
#include * * * ** ** *
#include **** * * **
int main(void){
* **** * * ******* * ** * * ** * * ** ** * * ** * *
** * * * ** *
** * * ** * * * ** * * * ** ** * * * *
* * * * ** ** * **
*** * * ** * * * * * ** ** * * *
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* * * *
* * ** * * *
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*** * * * * *** * ** * * **
}
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* * * **
* * * * *** * **** *
* * * *** * *** * ++;
* * * * * * *
}
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* * * * * ** * **** ***
* * ** ** (i = 0;i <= sizes;i++)
* * **
** * * * *
* * * * * * * = s[i];
* * * *
* * ** ***
*** * *** (i = 0;i <= sizet;i++)
* * **
* * ** ** * *
*** * * ** * * = t[i];
* * * ** ** *
** * * *
*** * * * * * * ** * ** * * ** * *
* * * * (i = 0;i <= sizes;i++)
{
** * ** * *
* * * ** * ** ** * *
* *** * * ** = p[i] - 32;
* * * * * *
**** * ** *
* * * ** ** ** = p[i];
}
* * * * * ** ** * ** *
for (i = 0;i <= sizet;i++)
{
** * *** * **
* * ** * * * * **
*** ** *** ** ** = f[i] - 32;
* * * * * ** **
* * * ***
* * * * * ** ** = f[i];
* * * ** *
}
* * * ** ** * * ** * *
//asplpSLP slp
** * ** (i = 0;p2[i] != '\0';i++)
* * * * *
** * ** * * (p2[i] == f2[0])
** ** * * * * ** ** *
** * * * ** * * ** *
* * ** **** * ** * ** = i;
* * * * **
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* * * * * ** * *
* * * * ** * * * ** * * * **
* ** * ** * * * * ** * * *
** * **** * ** * * *** * * * (p2[j+u] == f2[u])
* * * * * * ** *** * * *
* ** ** * ** * * * * * ** ** * ** * * += 1;
* * * * * ** * * * *
* * * *** * ** * * ** * **
** **** * * ** * * * * *** *
* ** ** * * * ** * * * * **
** * * * * * * * * * * * * ** *
* * * * * *** * * * * * * * * ** = 0;
** * * * ** * * * * *
** * * ** ** * * ** ** * * * * * **
** ** *** ** * ** * * * *
* *** ** * * *** ** * * ** ** ***
**** * *** ** ** ** ** * * *
* * * * ** * * * *
* ** ** * *** * ** ****
* * ** * *
* * ** * * **
** * *
* * ** * * * * ** * * * ** (d > 0)
** * * ** * *** * * ** * * *** * **
* * **** * * * * *** * ** ** ** * *** is * ** * ** ** in ** **** * ** * ** * * * *
* * * ** * ** ** * ** * * ** * * ***** * *
* * * * * ** ** **
* * * * * * ***
* ** * * ** * * * * ** * *** ** * *** are no * ** *** * *** in *** * * * * * *** **** * *
** ** * * * * * ** * * * 0;
* ** ** * ** * ** **
}
answered
Dec 10, 2016
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蘇瑋朝
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